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A 392 -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2}\) . Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J}\) . Calculate \(h\) .

Short Answer

Expert verified
The height \( h \) is approximately 11.73 meters.

Step by step solution

01

Calculate Initial Kinetic Energy

The initial kinetic energy consists of both translational and rotational kinetic energy. The translational kinetic energy \( KE_{trans} \) is given by \( \frac{1}{2} M v^2 \) and the rotational kinetic energy \( KE_{rot} \) is \( \frac{1}{2} I \omega^2 \). First, we need to express the velocity \( v \) in terms of \( \omega \) using \( v = \omega R \), where \( R = 0.600 \text{ m}\). Therefore, \( v = 0.600 \times 25.0 \).
02

Calculate Velocity and Substitute Back

Using the relation \( v = \omega R \), we calculate \( v = 0.600 \times 25.0 = 15.0 \text{ m/s} \). The mass \( M \) can be found using the weight \( W = M g \), so \( M = \frac{W}{g} = \frac{392}{9.8} = 40 \text{ kg} \). Substitute the value of \( v \) into the expressions for kinetic energy: \( KE_{trans} = \frac{1}{2} (40) (15.0)^2 \) and \( KE_{rot} = \frac{1}{2} (0.800 \times 40 \times 0.600^2) (25.0)^2 \).
03

Calculate Total Initial Kinetic Energy

Compute \( KE_{trans} = \frac{1}{2} \times 40 \times 225 = 4500 \text{ J} \). For \( KE_{rot} \), calculate the moment of inertia \( I = 0.800 \times 40 \times (0.600)^2 = 11.52 \text{ kg m}^2 \). Then compute \( KE_{rot} = \frac{1}{2} \times 11.52 \times 625 = 3600 \text{ J} \). The total initial kinetic energy is \( KE = 4500 + 3600 = 8100 \text{ J} \).
04

Apply Conservation of Energy

Using the work-energy principle, we know that the change in kinetic energy is equal to the work done by friction. Thus, \( KE_{initial} - KE_{final} = |W_{friction}| \). Since the wheel comes to stop at height \( h \), \( KE_{final} = 0 \). Therefore, \( 8100 - 0 = 3500 \). So at the top \( mgh = 8100 - 3500 \). Substitute the known values to solve for \( h \).
05

Solving for Height, h

The equation becomes \( mgh = 4600 \text{ J} \). Using the mass \( m = 40 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \), \( h = \frac{4600}{40 \times 9.8} = \frac{4600}{392} \approx 11.73 \text{ meters} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object has due to its motion. It comes in two forms in the realm of physics: translational and rotational. In the given problem, we analyze both types to fully understand the motion of the wheel.

**Translational Kinetic Energy**
  • This type of energy depends on the linear motion of an object's center of mass.
  • It is calculated using the formula \( KE_{trans} = \frac{1}{2} M v^2 \), where \( M \) is the mass and \( v \) is the velocity.
  • Here, the velocity \( v \) is related to the wheel's rotation by \( v = \omega R \), where \( \omega \) is the angular velocity and \( R \) is the radius.
**Rotational Kinetic Energy**
  • This energy arises from the rotation of an object around an axis.
  • It is given by \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia.
  • In our problem, both kinetic energies contribute to the total energy as the wheel moves.

By calculating these energies, we determine the wheel's total kinetic energy, influencing how far it travels up the hill before stopping.
Moment of Inertia
The moment of inertia \( I \) is a critical factor in rotational motion. It determines how an object resists rotation and is akin to mass in linear motion. For any given axis, the moment of inertia varies based on the distribution of mass around that axis.

**Determining Moment of Inertia**
  • In our scenario, the formula for the wheel’s moment of inertia is \( 0.800MR^2 \).
  • This formula reflects the fact that most of the wheel's mass is distributed around its edge, as typical for wheels and cylinders.
**Factors Influencing Moment of Inertia**
  • The more mass concentrated further from the axis, the larger the moment of inertia, making it harder to spin.
  • The radius also plays a significant role, as a larger radius results in an increased moment of inertia.

Understanding the moment of inertia allows us to correctly calculate the rotational kinetic energy and analyze the wheel’s behavior as it rolls.
Rotational Motion
Rotational motion describes an object's movement around a central point or axis. It is distinct yet interconnected with linear motion. To understand rotational dynamics, we consider quantities like angular velocity, angular acceleration, and torque. However, in this exercise, the focus lies primarily on angular velocity and the relationship to translational motion.

**Angular Velocity**
  • Symbolized by \( \omega \), angular velocity measures how swiftly an object rotates.
  • Units are typically radians per second (rad/s), a measure of the angle through which an object rotates in a given time.
**Connection to Translational Motion**
  • Through the equation \( v = \omega R \), rotational speed translates to linear speed.
  • This links the wheel's rolling motion to its path along the ground.

In this exercise, understanding rotational motion helps us calculate effectively the rolling wheel's initial kinetic energy. This analysis is vital as it affects how the wheel moves from the bottom to the top of the hill, taking into account energy loss due to friction.

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Most popular questions from this chapter

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) s, starting from rest?

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

In a spring gun, a spring of force constant 400 \(\mathrm{N} / \mathrm{m}\) is com- pressed 0.15 \(\mathrm{m}\) . When fired, 80.0\(\%\) of the elastic potential energy stored in the spring is eventually converted into the kinetic energy of a \(0.0590-\mathrm{kg}\) uniform ball that is rolling without slipping at the base of a ramp. The ball continues to roll without slipping up the ramp with 90.0\(\%\) of the kinetic energy at the bottom converted into an increase in gravitational potential energy at the instant it stops. (a) What is the speed of the ball's center of mass at the base of the ramp? (b) At this position, what is the speed of a point at the top of the ball? (c) At this position, what is the speed of a point at the bottom of the ball? (d) What maximum vertical height up the ramp does the ball move?

A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . (a) If the disk rolls up a \(30.0^{\circ}\) ramp, how far along the ramp will it move before it stops? (b) Explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

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